Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

Please welcome our new moderators!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

122 submissions , 103 unreviewed
3,497 questions , 1,172 unanswered
4,545 answers , 19,342 comments
1,470 users with positive rep
408 active unimported users
More ...

A question from Ticcati's red QFT textbook.

+ 4 like - 0 dislike
13 views

From Ticcati's textbook, he asks to show that from the axioms of position operator we get that: $$ \text{e}^{-ia\cdot P} |x\rangle = |x+a\rangle $$

where the axioms are: $$ X=X^{\dagger} $$

If $\Delta_a$ is a space translation, then $U(\Delta_a)^{\dagger} X U(\Delta_a) = X + a$, where $U$ is the representation of a unitary operator, we know that $\text{e}^{ia\cdot P} X \text{e}^{-ia\cdot P} = X + a$.

If $R$ is a space rotation then $U(R)^{\dagger} X U(R) = RX$.

Here's what I tried so far to do with this:

$$ \text{e}^{ia\cdot P} X \text{e}^{-ia\cdot P}|x+a\rangle =(X+a) |x+a\rangle = (x+a) |x+a\rangle $$

$$ X|y\rangle:=X \text{e}^{-ia\cdot P}|x+a\rangle = (x+a)\text{e}^{-ia\cdot P}|x+a\rangle=(x+a) |y\rangle $$

Now I want to show somehow that $|y\rangle=|x+2a\rangle$, but I don't see how, any hints?

Thanks in advance.

This post has been migrated from (A51.SE)
asked Dec 6, 2011 in Theoretical Physics by MathematicalPhysicis (10 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
You should post this as an answer so it can be accepted.

This post has been migrated from (A51.SE)
Ah, that's what I did, I wasn't sure it was right cause I thought to myself that this eigenvalue could belong to another eigenvector, but now I see that it doesn't matter as long as it belongs to $|x+a>$. Foolish me...

This post has been migrated from (A51.SE)
Just act by $X$ on $e^{-iaP}|x\rangle$: $Xe^{-iaP}|x\rangle=e^{-iaP}(X+a)|x\rangle=e^{-iaP}(x+a)|x\rangle=(x+a)e^{-iaP}|x\rangle$. I.e. $e^{-iaP}|x\rangle$ is an eigenvector with eigenvalue $x+a$.

This post has been migrated from (A51.SE)

1 Answer

+ 4 like - 0 dislike

As requested, reposting the comment.

Assuming one knows what an eigenvector for a continuous spectrum is (I don't), the solution is straightforward. Act by $X$ on $e^{-iaP}|x\rangle$: $$Xe^{-iaP}|x\rangle=e^{-iaP}(X+a)|x\rangle=e^{-iaP}(x+a)|x\rangle=(x+a)e^{-iaP}|x\rangle,$$ i.e. $e^{-iaP}|x\rangle$ is an eigenvector of $X$ with eigenvalue $x+a$. One of the axioms you are missing should state that the spectrum of $X$ is simple, which then implies that $e^{-iaP}|x\rangle\sim|x+a\rangle$. Normalization of $|x\rangle$ gives an equality.

Perhaps, this can be made precise using the machinery of Gelfand triples which I am unfamiliar with.

However, stating the condition of the simplicity of the continuous spectrum does not require anything fancy. The spectral theorem (for unbounded self-adjoint operators) allows us to view $X$ as a multiplication operator by a function $f:M\rightarrow\mathbf{R}$ on $L^2(M,\mu)$. The requisite condition is that $f$ is injective. Note, that it is automatically surjective unless $M$ is empty.

This post has been migrated from (A51.SE)
answered Dec 10, 2011 by Pavel Safronov (1,115 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...